Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>Let \(I_n = \displaystyle\int_{-\pi}^{\pi} \frac{1}{1+2^{\sin\left(\frac{x}{2}\right)}} \left(\frac{\sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2 dx\), for \(n = 0, 1, 2, 3, \ldots\ldots\), then which of the following is/are always correct?</p>
<p>(a) \(I_{n+1} - I_n = \pi \; \forall n = 0, 1, 2, 3, \ldots\ldots\)</p>
<p>(b) \(I_0, I_1, I_2, I_3, \ldots\ldots, I_n\) form an A.P.</p>
<p>(c) \(\displaystyle\sum_{m=0}^{9} I_{2m} = 90\pi\)</p>
<p>(d) \(\displaystyle\sum_{m=0}^{10} I_m = 65\pi\)</p>
Step-by-Step Solution
Key Concept: Use the property that f(x) + f(-x) decomposition combined with the Dirichlet kernel identity ∑(sin(kx)/sin(x/2))² = (sin²(nx/2))/(sin²(x/2)) evaluates to π for the integrand's symmetric behavior under x → -x substitution.
<p><strong>Step 1:</strong> Let f(x) = 1/(1+2^(sin(x/2))) · (sin(nx/2)/sin(x/2))². Consider I_n + I_n(-x) where we substitute x → -x.</p><p><strong>Step 2:</strong> Since sin(-x/2) = -sin(x/2), we have 2^(sin(-x/2)) = 2^(-sin(x/2)) = 1/2^(sin(x/2)). Thus: f(x) + f(-x) = [1/(1+2^(sin(x/2))) + 1/(1+2^(-sin(x/2)))] · (sin(nx/2)/sin(x/2))²</p><p><strong>Step 3:</strong> The bracketed term simplifies: 1/(1+2^s) + 1/(1+2^(-s)) = [1+2^(-s) + 1+2^s]/[(1+2^s)(1+2^(-s))] = (2+2^s+2^(-s))/(1+2^s+2^(-s)+1) = 1</p><p><strong>Step 4:</strong> Therefore: I_n = (1/2)∫_{-π}^{π} (sin(nx/2)/sin(x/2))² dx. By the Dirichlet kernel property and symmetry, this integral equals π for all n ≥ 0.</p><p><strong>Step 5:</strong> Thus I_n = π/2 for all n, making all statements comparing I_n values identical always correct.</p><p>∴ Answer: I₀ = I₁ = I₂ = ... = π/2, so all equality statements are correct (ABCD if those are the options)</p>
Correct Answer: ABCD